JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 3)
Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} $ and $ \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} $. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:
$\sqrt{10}$
$2$
$2\sqrt{3}$
$3$
Explanation
Vector parallel to 'L'
$$\begin{aligned} & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{array}\right|=10 \hat{i}-10 \hat{j}+5 \hat{k} \\ & =5(2 \hat{i}-2 \hat{j}+\hat{k}) \end{aligned}$$
Equation of 'L'
$$\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-3}{1}=\lambda(\text { say })$$
Let $\mathrm{Q}(2 \lambda+2,-2 \lambda-1, \lambda+3)$
$$\begin{aligned} & \Rightarrow 2 \lambda+2=0 \Rightarrow \lambda=-1 \\ & \Rightarrow \mathrm{Q}(0,1,2) \\ & \mathrm{d}(\mathrm{P}, \mathrm{Q})=3 \end{aligned}$$
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